# The rectangular coordinates of a point are given. Plot the point and find two sets of polar coordinates for the point for `0 <=thetalt= 2pi` .(-3,-3) please help not really sure what to do

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### 1 Answer

A point in (x,y) coordinates can be given in polar coordinates as below,

`(x,y) = (rcos(theta),rsin(theta))`

where r is `sqrt(x^2+y^2)` and theta is the angle with the x-axis.

The point in this example is (-3,-3). This in the third quadrant.

`r = sqrt((-3)^2+(-3)^2)`

`r = sqrt(9+9)`

`r = 3sqrt(2)`

`x = rcos(theta)`

`-3=3sqrt(2)cos(theta)`

`cos(theta) = -1/sqrt(2)`

Since the point is in the third quadrant, `theta = pi+ pi/4`

`theta =(5pi)/4`

If you solve for y also, you will get the same, `theta = (5pi)/4`

Therefore,

`(-3,-3) = (3sqrt(2)cos((5pi)/4), 3sqrt(2)sin((5pi)/4))` in polar coordinates.

We can write this as,

`(-3,-3) = (3sqrt(2)cos(pi+pi/4), 3sqrt(2)sin(pi+pi/4))`

cos(pi+pi/4) = -cos(pi/4) and sin(pi+pi/4) = -sin(pi/4)

Now,

`(-3,-3) = (-3sqrt(2)cos(pi/4),-3sqrt(2)sin(pi/4))`

Therefore two sets of polar coordiantes are,

`(-3,-3) = (3sqrt(2)cos(pi+pi/4), 3sqrt(2)sin(pi+pi/4))`

and

`(-3,-3) = (-3sqrt(2)cos(pi/4),-3sqrt(2)sin(pi/4))`